
now that we know what are the x-values, what are the y-values? well, we can just use the 2nd equation, since we know that y = x - 28, then
![\bf y = x - 28\implies \stackrel{\textit{when x = 16}}{y = 16 - 28}\implies y = -12 \\\\\\ y = x - 28\implies \stackrel{\textit{when x = 12}}{y = 12 - 28}\implies y = -16 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{x}{16}~~,~~\stackrel{y}{-12})\qquad,\qquad (\stackrel{x}{12}~~,~~\stackrel{y}{-16})~\hfill](https://tex.z-dn.net/?f=%5Cbf%20y%20%3D%20x%20-%2028%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwhen%20x%20%3D%2016%7D%7D%7By%20%3D%2016%20-%2028%7D%5Cimplies%20y%20%3D%20-12%20%5C%5C%5C%5C%5C%5C%20y%20%3D%20x%20-%2028%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwhen%20x%20%3D%2012%7D%7D%7By%20%3D%2012%20-%2028%7D%5Cimplies%20y%20%3D%20-16%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%28%5Cstackrel%7Bx%7D%7B16%7D~~%2C~~%5Cstackrel%7By%7D%7B-12%7D%29%5Cqquad%2C%5Cqquad%20%28%5Cstackrel%7Bx%7D%7B12%7D~~%2C~~%5Cstackrel%7By%7D%7B-16%7D%29~%5Chfill)
Answer:
15√15−15√5+15√3−15
Step-by-step explanation:
Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.
Exact Form: 15√15−15√5+15√3−15
Decimal Form: 35.53449264…
Step-by-step explanation:

Answer:
A. 6
Step-by-step explanation:
Using the Pythagorean theorem which states that: <u>Hypotenus² = Opposite² + Adjacent²</u>
Where: hypotenus = 10, opposite = x, adjacent = 8
So:

Solving for x

Collect like terms to make x the subject of formula


square root both sides of the equation to find the value of x

Therefore: Option A is correct